Problem:
In triangle ABC, the medians AD and CE have lengths 18 and$ 27 , respectively, and AB=24. Extend CE to intersect the circumcircle of ABC at F. The area of triangle AFB is mn, where m and n are positive integers and n is not divisible by the square of any prime. Find m+n.
Solution:
Let P be the intersection of AD and CE. Since angles ABF and ACF intercept the same arc, they are congruent, and therefore triangles ACE and FBE are similar. Thus EF/12=12/27, yielding EF=16/3. The area of triangle AFB is twice that of triangle AEF, and the ratio of the area of triangle AEF to that of triangle AEP is 916/3, since the medians of a triangle trisect each other. Triangle AEP is isosceles, so the altitude to base PE has length 122−(9/2)2=(1/2)242−92=(3/2)82−32=(3/2)55, and the area of triangle AEP is (27/4)55. Therefore, [AFB]=2[AFE]=2(16/27)⋅[AEP]=2(16/27)⋅(27/4)55=855, and m+n=63.
